Summary for aluminum circuit board trace Discussion
Aluminum circuit board trace sizing is the core issue here: Gangadharan Sivaraman asked how thick a 400 micrometer-wide aluminum deposition must be to carry 3.5 amps, but noted that the length of the run was unknown. With no answer text provided in the thread excerpt, the practical takeaway is that the required thickness cannot be calculated from width and current alone; the trace length, allowable voltage drop, resistivity, temperature rise, and cooling conditions all matter. In other words, the question is underdetermined as stated, so any reliable design would need more electrical and thermal details before a thickness recommendation can be made.
Q. I have to deposit Aluminum and I know its width is 400 micrometer and I want to calculate the thickness of the deposition so that the Aluminum can carry 3.5 Amps and I have no details about the length of the deposition.
Please help me out.
Gangadharan Sivaraman
Texas Tech University – Lubbock, Texas
2002
 | A. I am 100 % positive that you can find data on the load capacity of aluminum in the TT Library. James Watts A. I assume you are designing a circuit! Your concern needs to be current density and not length. Your spec of 3.5 A is a very high current to be carried by a 0.4 mm Al strip. If you don’t use enough thickness, you will experience “electromigration”. Check some literature on that subject. A lot has been studied and published in the context of ICs and hybrids. Having suggested that, there are some rules of thumb in the design. Determine the thickness based on the current density not to exceed 1 megamp per cm sq. or less. Then some (50% to 100%) safety factor for coating defects. As a good start, I would consider 10E5 density and find cross section, then determine thickness based on density of Al. Thicker the coating, better off you will be. Then worry about process economics. Mandar Sunthankar A. In the macro world we use about 600 Amps per square inch of cross section as a good safe rating for aluminum. That probably has very little to do with the answer you need for your micro device, but there it is for reference 🙂
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Q. Dear James Watts, Mandar Sunthankar, and Ted Mooney,
I would like to Thank you for your valuable comments and I have used the current density to find the thickness and I am enclosing the solution with this.
I request you to review this and comment on any possible mistakes in this calculations.
Thanking you Very much
Aluminum is 400 Microns wide, how thick it has to be to support
3.5 Amps?
Width of Aluminum is 400
Required Current I = 3.5 Amps
Operating voltage 3.3 V D.C
I assume J ( Current Density ) to be 10 e 5 A/cm2 = 1 milli Amps per Micro Meter Square
I = 3.5 A
J = I / A
A = I / J
3.5 / 1 10 -3
A =3500 Micrometer Square
A = W * T
W = 400 Microns
So T = 3500/400 = 8.75 Micrometer ( Micron )
Gangadharan Sivaraman [returning]
– Lubbock, Texas
A. Hi, Gangadharan. Your conversion factors and math appear correct.
So, assuming the allowable current density is 100,000 A/cm2 as your calculation asserts, then 8.75 micrometer thickness would be the correct answer.
But I don’t think that your allowable current density can possibly be correct. I don’t have experience in this field, but it sounds a good 10X too high to me, even for copper let alone aluminum, based on some quick reading.
Regards,
Ted Mooney, P.E.
Striving to live Aloha
finishing.com – Pine Beach, New Jersey
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